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Chapter 4 Interaction with External Sources 4.1 Classical Fields and Green’s Functions We start our discussion of sources with classical (non-quantized) fields. Consider L = 12 (@µ (x))2 − 12 m2 (x)2 + J(x) (x) Here J(x) is a non-dynamical field, the source. The equation of motion for the dynamical field ( (x)) is (@ 2 + m2 ) (x) = J(x) . (4.1) The terminology comes from the more familiar case in electromagnetism. The non-homogeneous Maxwell equations, � = ⇢, � ⋅E ∇ � − @0 E � = J� � ×B ∇ have as sources of the electric and magnetic fields the charge and current densities, � respectively. We can go a little further in pushing the analogy. In terms ⇢ and J, � respectively, the fields are of the electric and vector potentials, A0 and A, � = −@0 A� − ∇A � 0, E so that Gauss’s law becomes �=∇ � × A� B � ⋅ A� − ∇2 A0 = ⇢ −@0 ∇ � ⋅ A� + @0 A0 = 0 this takes the form In Lorentz gauge, ∇ @ 2 A0 = ⇢ . 56 4.1. CLASSICAL FIELDS AND GREEN’S FUNCTIONS 57 This is exactly for the form of (4.1), for a massless field with the identification A0 for and ⇢ for J. In Lorentz gauge the equations satisfied by the vector potential are again of this form, @ 2 A� = J� . The four vector J µ = (⇢, J i ) serves as source of the four-vector potential Aµ . Writing @ 2 Aµ = J µ . is reassuringly covariant under Lorentz transformations, as it should be since this is where relativity was discovered! Each of the four components of Aµ satisfies the massless KG equation with source. Consider turning on and o↵ a localized source. The source is “on,” that is non-vanishing, only for −T < t < T . Before J is turned on is evolving as a free KG field (meaning free of sources or interactions), so we can think of the initial conditions as giving (x) = in (x) for t < −T , with in (x) a solution of the free KG equation. We similarly have that for t > T (x) = out (x) where out (x) is a solution of the free KG equation. Both in (x) and out (x) are solutions of the KG equations for all t but they agree with only for t < −T and t > T , respectively. Our task is to find out (x) given in (x) (and of course the source J(x)). Solve (4.1) using Green functions: (x) = hom (x) + � where the Green function satisfies d4 y G(x − y)J(y) (@ 2 + m2 )G(x) = (4) (x) (4.2) (4.3) and hom (x) is a solution of the associated homogeneous equation, which is the KG equation, (@ 2 + m2 ) hom (x) = 0 . We can use the freedom in hom (x) to satisfy boundary conditions. We can determine the Green function by Fourier transform, Then G(x) = � (@ 2 + m2 ) � so that d4 k ik⋅x ̃ e G(k) (2⇡)4 (4) (x) = � d4 k ik⋅x e (2⇡)4 d4 k ik⋅x ̃ d4 k ik⋅x d4 k ik⋅x 2 2 ̃ e G(k) = e (−k + m ) G(k) = e � � (2⇡)4 (2⇡)4 (2⇡)4 ̃ G(k) =− k2 1 . − m2 58 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES So preliminarily take G(x) = − � d4 k eik⋅x (2⇡)4 k 2 − m2 However, note this is ill-defined � since the integrand diverges at k 2 = m2 . The integral over k 0 diverges at k 0 = ± k�2 + m2 , or k 0 = ±Ek� for short: 0 � dk 0 eik t . (k 0 − Ek� )(k 0 + Ek� ) This integral can be thought of as an integral over a complex variable z along a contour on the real axis, Re(z) = k 0 , from −∞ to ∞. Then the points z = ±Ek� are locations of simple poles of the integrand, and we can define the integral by deforming the contour to go either above or below these poles. For example we can take the following contour: Im(z) × −Ek� × Ek� k 0 = Re(z) Regardless of which deformation of the contour we choose, the contour can be closed with a semicircle of infinite radius centered at the origin on the upper half-plane if t > 0 and in the lower half-plane if t < 0: Im(z) t>0 × −Ek� Im(z) t<0 × Ek� k 0 = Re(z) × −Ek� × Ek� k 0 = Re(z) 4.1. CLASSICAL FIELDS AND GREEN’S FUNCTIONS 59 because he integral along the big semicircle vanishes as the radius of the circle is taken infinitely large. For the choice of contour in this figure no poles are enclosed for t > 0 so the integral vanishes. This defined the advanced Green’s function, Gadv (x) = 0 for t > 0. Alternatively we can “displace” the poles by an infinitesimal amount −i✏, with ✏ > 0, so they lie just below the real axis, Im(z) Then we have × −Ek� − i✏ × k 0 = Re(z) Ek� − i✏ Gadv (x) = − � d4 k ik⋅x 1 e 4 (2⇡) (k 0 + i✏)2 − k� 2 − m2 Gret (x) = − � d4 k ik⋅x 1 e 4 0 2 (2⇡) (k − i✏) − k� 2 − m2 Similarly, the retarded Green’s function is with the contour bellow the two poles. It has Gret (x) = 0 for t < 0. For later use it is convenient to write Gret (x) = −✓(x0 ) � d3 k −ik� ⋅�x 1 �eiEk� t − e−iEk� t �� e �2⇡i (2⇡)4 2Ek� (4.4) Gadv (x) = ✓(−x0 ) � d3 k −ik� ⋅�x 1 �eiEk� t − e−iEk� t �� e �2⇡i (2⇡)4 2Ek� (4.5) = −i✓(x0 ) � (dk) (eik⋅x − e−ik⋅x ) = i✓(−x0 ) � (dk) (eik⋅x − e−ik⋅x ) We also note that one may choose a contour that goes above one pole and below the other. For example, going below −Ek� and above +Ek� we have GF (x) = − � = −� d4 k ik⋅x 1 e 4 0 (2⇡) (k − Ek� + i✏)(k 0 + Ek� − i✏) d4 k ik⋅x 1 e 4 2 (2⇡) (k − m2 + i✏) = i �✓(x0 ) � (dk) e−ik⋅x + ✓(−x0 ) � (dk) eik⋅x � (4.6) 60 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES In the solution (4.2), for x0 < −T the integral over y 0 only has contributions from x0 − y 0 < 0, and for x0 > T it has contributions only from x0 − y 0 > 0. So we have (x) = = Hence we can write out (x) = = in (x) + � out (x) + � in (x) + � in (x) + � d4 y Gret (x − y)J(y) d4 y Gadv (x − y)J(y) d4 y �Gret (x − y) − Gadv (x − y)�J(y) d4 y G(−) (x − y)J(y) (4.7) G(−) can be obtained from the di↵erence of (4.4) and (4.5). But more directly we note that × × − × × × × + × × is the same as The straight segments cancel and we are left with × This gives G(−) (x) = − � × d3 k −ik� ⋅�x 1 �eiEk� t − e−iEk� t �� e �2⇡i (2⇡)4 2Ek� � � = −i � (dk) �eiEk� t−ik ⋅�x − e−iEk� t−ik ⋅�x � d4 k ✓(k 0 ) (k 2 − m2 ) �eik⋅x − e−ik⋅x � (2⇡)3 d4 k = −i � "(k 0 ) (k 2 − m2 )eik⋅x (2⇡)3 = −i � where � � �+1 "(k ) = � � � �−1 0 k0 > 0 k0 < 0 4.2. QUANTUM FIELDS 61 You may have noticed that this looks a lot like the function + (x) = [ In fact, computing the commutator of free fields at arbitrary times, (+) (x), [ (x), (y)] = � (dk)(dk ′ )[↵� e−ik⋅x + ↵†� eik⋅x , ↵� ′ e−ik ⋅y + ↵†� ′ eik ⋅y ] k k = � (dk)�eik⋅(y−x) − e−ik⋅(y−x) � 4.2 k ′ k ′ (−) (0)]. = iG(−) (y − x) = −iG(−) (x − y) Quantum Fields We now consider equation (4.1) for the KG field with a source in the case that the field is an operator on the Hilbert space F. The source J(x) is a a c-number, a classical source, and still take it to be localized in space-time. Suppose the system is in some state initially, well before the source is turned on. Let’s take the vacuum state for definiteness, although any other state would be just as good. As the system evolves, the source is turned on and then o↵, and we end up with the system in some final state. Now, in the classical case if we start from nothing and turn the source on and then o↵ radiation is produced, emitted out from the region of the localized source. In the quantum system we therefore expect we will get a final state with any number of particles emitted from the localized source region. Then (x) = in (x) is the statement that for t < −T is the same operator as a solution to the free KG equation. Likewise for = out for t > T . Does this mean out = in ? Surely not, it is not true in the classical case. Since satisfies the equal time commutation relations, i[@t (x), (y)] = (3) (� x − y� ), etc, so do and . Since the latter solve the free KG equation, they have expansions in out in creation and annihilation operators, and the same Hilbert space. More precisely, † † we can construct a Fock space Fin out of ↵in , and a space Fout out of ↵out . These spaces are just Hilbert spaces of the free KG equation, and therefore they are isomorphic, Fin ≈ Fout ≈ FKG . So there is some linear, invertible operator S ∶ FKG → FKG , so that � �out = S −1 � �in . Since the states are normalized, S must preserve normalization, so S is unitary, S † S = SS † = 1. The operator S is called the S-matrix. Let’s understand the meaning of � �out = S † � �in . The state of the system initially (far past) is � �in . It evolves into a state � �out = S † � �in at late times, well after the source is turned o↵. We can expand it in a basis of the Fock space, the states �k� 1 , . . . , k� n � for n = 0, 1, 2, . . . In particular, if the initial state is the vacuum, then the expansion of S † �0� in the Fock space basis tells us the probability amplitude for emitting any number of particles. More generally out � � �in = in � �S� �in 62 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES is the probability amplitude for starting in the state � � and ending in the state � �. Note that the left hand side is not to be taken literally as an inner product in the same Hilbert space of free particles (else it would vanish except when the initial and final states are physically identical, i.e., when nothing happens). Now, S † in (x)S is an operator acting on out states that satisfies the KG equation. Similarly, if ⇡in (x) = @t in (x), S † ⇡in (x)S acts on out states. Moreover, the commutator [S † in (x)S, S † ⇡in (y)S = S † [ in (x), ⇡in (y)]S = [ in (x), ⇡in (y)] since the commutator is a c-number and S † S = 1, and similarly for the other commutators of S † in (x)S and S † ⇡in (x)S. This means that up to a canonical transformation, out (x) = S† in (x)S . Since we are free to choose the out fields in the class of canonical equivalent fields, we take the above relation to be our choice. There is a simple way to determine S. We will present this now, but the method works only for the case of an external source and cannot be generalized to the case of interacting fields. So after presenting this method we will present a more powerful technique that can be generalized. From Eq. (4.7) we have S† Recall in (x)S = = in (x) + � in (x) + i � d4 y G(−) (x − y)J(y) d4 y [ in (x), in (y)J(y)] . eA Be−A = B + [A, B] + 12 [A, [A, B]] + � = B + [A, B] It follows that S = exp �i � d4 y It is convenient to normal-order S. Since [ use if [A, [A, B]] = 0. in (y)J(y)� (+) (x), eA eB = eA+B+ 2 [A,B] 1 (4.8) (−) (y)] is a c-number we can which is valid provided [A, B] commutes with both A and B. So using i ∫ d4 x and i ∫ d4 x (+) (x)J(x) for A and B, S = ei ∫ d4 x (−) (x)J(x) ei ∫ d4 x (+) (x)J(x) e2 ∫ 1 d4 x d4 y [ (−) (x), (+) (y)]J(x)J(y) (−) (x)J(x) 4.2. QUANTUM FIELDS From (2.10), + (x2 63 − x1 ) ≡ [ ˆ(+) (x1 ), ˆ(−) (x2 )] = � (dk) e−ik⋅(x2 −x1 ) =� d4 k ✓(k 0 ) (k 2 − m2 )e−ik⋅(x2 −x1 ) (2⇡)4 and introducing the Fourier transform of the source, J(x) = � we have 4 4 � d xd y[ (−) in (x), d4 k ik⋅x ̃ e J(k) (2⇡)4 (+) in (y)]J(x)J(y) d4 ki ̃ 2 )J(k ̃ 3 ) � d4 x � d4 y eik2 ⋅x eik3 ⋅y eik1 ⋅(y−x) ✓(k10 ) (k12 − m2 )J(k 4 (2⇡) i=1 3 = −� � = −� d4 k 2 ̃ J(−k) ̃ ̃ ✓(k 0 ) (k 2 − m2 )J(k) = − � (dk)�J(k)� , (2⇡)4 ̃ where we have used J ∗ (x) = J(x) ⇒ J(−k) = J̃∗ (k), and it is understood that 0 k = Ek� . Hence, S = ei ∫ d4 x (−) (x)J(x) ei ∫ d4 x (+) (x)J(x) ̃ e− 2 ∫ (dk)�J(k)� . 1 2 As an example of an application, we can compute the probability of finding particles in the final state if we start form no particles initially (emission probability). Start from probability of persistence of the vacuum, 2 ̃ �out �0�0�in �2 = �in �0�S�0�in �2 = exp �− � (dk)�J(k)� �. Next compute the probability that one particle is produced with momentum k� : �out �k� �0�in �2 = �in �k� �S�0�in �2 = �in �0�↵k� S�0�in �2 . where “in” in ↵k� is implicit. To proceed we use so that e−i ∫ (−) J Hence [↵k� , ↵k� ei ∫ (−) (x)] = � (dk ′ )[↵k� , ↵†� ′ ]eik ⋅x = eik⋅x (−) J k = ↵k� − i � [ (−) ′ (with k 0 = Ek� ) ̃ , ↵k� ]J = ↵k� + i � d4 xeik⋅x J(x) = ↵k� + iJ(−k) . 2 2 ̃ ̃ �out �k� �0�in �2 = �J(k)� exp �− � (dk)�J(k)� �. 64 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES 4.2.1 Phase Space Suppose we want to find the probability of finding a single particle as t → ∞ regardless of its momentum, that is, in any state. We need to sum over all final states consisting of a single particle, once each. Since we have a continuum of states we have to integrate over all k� with some measure, ∫ dµ(k� ). Clearly dµ(k� ) must involve d3 k, possibly weighted by a function f (k� ). Presumably this function is rotationally invariant, f = f (�k� �). But we suspect also dµ(k� ) is Lorentz invariant, so dµ(k� ) ∝ (dk), that is, it equals the invariant measure up to a constant. Let’s figure this out by counting. Note that how we normalize states matters. First we determine this via a shortcut, and later we repeat the calculation via a more physical approach (and obtain, of course the same result). We want 2 � �out �n�0�in � = � in �0�n�out out �n�0�in = out �0� �� �n�out out �n�out � �0�in , n n n where the sum is restricted over some states. The operator � �n�out out �n� n is a projection operator onto the space of “some states,” the ones we want to sum in the final state. We have already discussed this projection operator for the case of one particle states when �k� � is relativistically normalized. It is � (dk)�k� ��k� � . So we have 2 2 ̃ ̃ 1-particle emission probability = � (dk)�J(k)� exp �− � (dk)�J(k)� �. Now we repeat the calculation by counting states. It is easier to count discrete sets of states. We can do so by placing the system in a box of volume Lx Ly Lz . � Take, say, periodic boundary conditions. Then instead of ∫ (dk)(↵k� eik ⋅�x + h.c.), we � have a Fourier sum, ∑k� (ak� eik ⋅�x + h.c.). Here ak� are creation operators with some normalization we will have to sort out. For periodic boundary conditions we must have kx Lx = 2⇡nx , ky Ly = 2⇡ny , kz Lz = 2⇡nz , with ni integers. We label the one n particle states by these, �� n � = a†� �0�, where k� = 2⇡ � Lnxx , Lyy , Lnzz �. The probability of finding � � in state �� n � is ��� n � ��2 if both � � and �� n � are normalized to unity. † (Note that this means [ak� , a� ′ ] = nx n′x ny n′y nz n′z ). The probability of finding � � k in a 1-particle state is then ∑n� ��� n � ��2 . Let’s be more specific. What is the probability of finding � � in a 1-particle state with momentum in a box (kx , kx + kx ) × (ky , ky + ky ) × (kz , kz + kz )? k 4.2. QUANTUM FIELDS There are nx ny nz = 65 Lx Ly Lz (2⇡)3 kx ky kz state in the box. For large volume the spacing between values of k� becomes small, so in the limit of large volume kx → dkx , etc, and the number of particles in the momentum box is kx ky kz → V Lx Ly Lz (2⇡)3 d3 k (2⇡)3 where V = Lx Ly Lz is the volume fo the box. As we switch from discrete to continuum labels for our states we must be careful with their normalization condition. Since on the space of 1-particle states we have n ��� n� = 1 � �� � n ⇒ n ��� n �� n ′ � = �� n ′� � �� � n then in the limit, denoting by �k� � the continuum normalized states, � � n V 3k �� n ��� n �� n ′ � = �� n ′ � → � d3 k�k� ��k� �k� ′ � = �k� ′ � (2⇡)3 V ⇒ n n′ (2⇡)3 x x ny n′y nz n′z → (3) (k� − k� ′ ) . Putting these elements together, the probability of finding � � in a 1-particle state is n � ��2 → � d3 k��k� � ��2 . � ��� � n Finally, if we want to change the normalization of states so that �k� ′ �k� � = Nk� k� ′ ) then the probability of finding � � in a 1-particle state is (3) (k� − d3 k � ��k � ��2 . � Nk� For relativistic normalization of states, Nk� = (2⇡)3 2Ek� and the probability is 2 � (dk)��k� � �� . This is precisely our earlier result. Denoting the emission probability for n particles in the presence of a localized source J(x) by pn , we have p0 = e−⇠ , p1 = ⇠e−xi , ̃ k� )�2 . where ⇠ = � (dk)�J( (4.9) 66 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES 4.2.2 Poisson We can now go on to compute pn for arbitrary n. Start with n = 2, that is, emission of two particles. e−i ∫ (−) J ↵k� ↵k� ′ ei ∫ (−) J so that = e−i ∫ (−) J ↵k� ei ∫ (−) J e−i ∫ (−) J ↵k� ′ ei ∫ (−) J ̃ ̃ ′ = (↵k� + iJ(−k))(↵ � ′ + iJ(−k )) (4.10) k � � ′ �0�in = −J(−k) ̃ ̃ ′ )e− 12 ⇠ . J(−k out �k k To get the emission probability we must sum over all distinguishable 2-particle states. Since �k� k� ′ � = �k� ′ k� � we must not double count. When we integrate over a box in momentum space, we count twice the state �k� k� ′ � if we sum over k1 and k2 with values k1 = k, k2 = k ′ and k1 = k ′ , k2 = k: kx′ = kx Hence 1 2 × kx′ kx p2 = 12 ⇠ 2 e−⇠ . The generalization is straightforward: pn = 1 n ⇠ exp(−⇠) . n! This is a Poisson distribution! Note that ∑n pn = 1, that is, there is certainty of finding anything (that is, either no or some particles). The mean of the distribution ̃ k� )�2 . is ⇠ = ∫ (dk)�J( 4.3 Evolution Operator We now introduce a more general formalism to derive the same results, but that will be more useful when we consider interacting quantum fields. We want to construct an operator U (t) that gives the connection between the field and the “in” field in : (� x , t) = U −1 (t) in (� x , t)U (t) . (4.11) Since we are assuming → in as t → −∞, we must have U (t) → 1 as t → −∞ . (4.12) 4.3. EVOLUTION OPERATOR The S matrix is then 67 S = lim U (t) . t→∞ The time evolution of and @ (� x , t) = i[H(t), (� x , t)] @t in are given by @ in (� x , t) = i[H0in (t), @t x , t)] in (� (4.13) Here H = H0 + H ′ , where H0 is the free Hamiltonian and H ′ describes interactions. In the present case, � )2 + 1 m2 H0 = � d3 x � 12 ⇡ 2 + 12 (∇ 2 2 � H ′ (t) = � d3 x J(� x , t) (� x , t) . Also the subscript “in” means the argument has “in” fields. So H = H( , ⇡, J) with H0 = H0 ( , ⇡) while H0in = H0in ( in , ⇡in ). From Eq. (4.11) we have U −1 (t)H( (t), ⇡(t), J(t))U (t) = H(U −1 (t) (t)U (t), U −1 (t)⇡(t)U (t), U −1 (t)J(t)U (t)) = H( and in (t), ⇡in (t), J(t)) (4.14) @ in @ �U (� (� x , t) = x , t)U −1 (t)� @t @t dU dU −1 = U −1 + U + U i[H(t), (� x , t)]U −1 dt dt dU −1 dU −1 = U U + i[Hin (t), in (� x , t)] in − in dt dt dU −1 = i[−i U + Hin (t), in (� x , t)] dt Comparing with Eq. (4.13) this is the commutator with H0in so we must have or −i dU −1 U + Hin (t) = H0in (t) dt dU ′ = −i(Hin − H0in )U = −iHin U (4.15) dt The solution to this equation with the boundary condition (4.12) gives the S matrix, S = U (∞). Note that the equation contains only “in” fields, which we know how to handle. We can solve (4.15) by iteration. Integrating (4.15) form −∞ to t we have U (t) − 1 = −i � t −∞ ′ dt′ Hin (t′ )U (t′ ) . 68 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES Now use this again, repeatedly, U (t) = 1 − i � ′ dt′ Hin (t′ ) �1 − i � t −∞ = 1 − i� t′ −∞ ′ dt1 Hin (t1 ) + (−i)2 � t −∞ ∞ = � = 1 + � (−i)n � t −∞ n=1 dt1 � t1 −∞ ′ dt′′ Hin (t′′ )U (t′′ )� t dt1 � −∞ dt2 � � t1 −∞ tn−1 −∞ ′ ′ dt2 Hin (t1 )Hin (t2 )U (t2 ) ′ ′ ′ dtn Hin (t1 )Hin (t2 )�Hin (tn ) . ′ ′ ′ Note that the product Hin (t1 )Hin (t2 )�Hin (tn ) is time-ordered, that is, the Hamiltonians appear ordered by t ≥ t1 ≥ t2 � ≥ tn . This is the main result of this section. We can write the result more compactly. Note that � t −∞ dt1 � t1 −∞ ′ ′ dt2 Hin (t1 )Hin (t2 ) = � t −∞ dt2 � t2 −∞ ′ ′ dt1 Hin (t2 )Hin (t1 ) The two integrals cover the whole t1 vs t2 plane, as can be seen from the followin figures in which the shaded regions correspond to the region of integration of teh first and second integrals, respectively: t2 t2 t1 t1 For any two time dependent operators, A(t) and B(t), we define the time-ordered product � � �A(t1 )B(t2 ) t1 > t2 T (A(t1 )B(t2 )) = ✓(t1 −t2 )A(t1 )B(t2 )+✓(t2 −t1 )B(t2 )A(t1 ) = � � � �B(t2 )A(t1 ) t2 > t1 and similarly when there are more than two operators in the product. Then ′ ′ dt2 Hin (t1 )Hin (t2 ) = t 1 t ′ ′ dt 1 � � � dt2 T �Hin (t1 )Hin (t2 )� . 2 −∞ −∞ −∞ −∞ For the term with n integrals there are n! orderings of t1 , . . . , tn so we obtain t dt1 � t1 t t t (−i)n ′ ′ ′ � dt � dt2 � � dtn T (Hin (t1 )Hin (t2 )�Hin (tn )) . (4.16) −∞ −∞ −∞ n=1 n! ∞ U (t) = 1 + � or, comparing with ez = ∑∞ n=0 1 n n! z we write U (t) = T �exp �−i � t −∞ ′ dt′ Hin (t′ )�� . 4.4. WICK’S THEOREM 69 You should keep in mind that the meaning of this expression is the explicit expansion in (4.16). Finally, taking t → ∞, S = T �exp �−i � or, using L′ = −H′ , ∞ −∞ ′ ′ dt Hin (t)�� = T �exp �−i � d4 x Hin �� S = T �exp �i � d4 x L′in �� . Let’s use this general result for the specific example we have been discussing, and compare with previous results. Use L′in = J(x) in (x). Then But we had obtained S = T �exp �i � d4 x J(x) in (x)�� S = e− 2 ⇠ ∶exp �i � d4 x J(x) 1 . in (x)�∶ (4.17) 2 ̃ with ⇠ = 12 ∫ (dk)�J(k)� , as in (4.9). Are these two expression for S the same? To show they are we need some additional machinery, some relation between T-ordered and normal-ordered products. 4.4 Wick’s Theorem In order to answer the question, what is the di↵erence between T-ordered and normal-ordered products, we consider T ( (x1 ) (x2 )) − ∶ (x1 ) (x2 )∶ Here and below stands for an “in” field, that is, a free field satisfying the KG equation. For notational conciseness we write i for (x1 , etc. Letting = (+) + (−) and taking x01 > x02 the di↵erence is ( (+) (−) (+) (−) (+) (+) (−) (+) (−) (+) (−) (−) 1 + 1 )( 2 + 2 ) − ( 1 2 + 1 2 + 2 1 + 1 + 2 ) =[ (+) 1 , (−) 2 ] This is a c-number (equals + (x2 − x1 )). Taling the expectation value in the vacuum we obtain, restoring full notation momentarily, T ( (x1 ) (x2 )) = ∶ (x1 ) (x2 )∶ +�0�T ( (x1 ) (x2 ))�0� . Clearly this holds for the more general case of several fields, T( n (x1 ) m (x2 )) =∶ ∶ n (x1 ) m (x2 )∶ +�0�T ( n (x1 ) m (x2 ))�0� . 70 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES Consider next the case of three fields, T ( take x01 ≥ x02 ≥ x03 : T( Now, 1 ∶ 2 3 ∶ = ( the (−) operators: 1 2 3) (+) 1 + (+) 1 ∶ 2 3∶ =∶ = 1 2 3 (−) 1 ) ∶ 2 3 ∶. (+) 1 2 3∶ 1 2 3) =∶ Without loss of generality 1 (∶ 2 3 ∶ +�0�T ( 2 3 )�0�) We need to move +[ where we have used the fact that [ �0�T ( 1 n )�0�. So we have T( = 1 2 3 ). (+) 1 , (+) 1 , (+) 1 (−) (+) 2 ] ∶ 3 ∶ +[ 1 , (−) n ] to the right of all of (−) 3 ] ∶ 2∶ is a c-number. In fact, it equals 1 2 3 ∶ + ∶ 1 ∶�0�T ( 2 3 )�0� + ∶ 2 ∶�0�T ( 1 3 )�0� + ∶ 3 ∶�0�T ( 1 2 )�0� . Of course, ∶ ∶ = , but the notation will more easily generalize below. More notation, rewrite the above as T( where 1 2 3) =∶ 1 2 3∶ + ∶ 1 2 3∶ + ∶ 1 2 3∶ + ∶ 1 2 3∶ n m is called a contraction. Wick’s theorem states that T( 1� n) =∶ 1� n∶ + + � ∶ pairs (i,j) � = �0�T ( . n m )�0� 1� i� j � n∶ 2-pairs (i,j),(k,l) ∶ 1 � i � j � k � l � n ∶ +� � � � n = even �∶ 1 2 3 4 � n−1 n ∶ +all pairings +� (4.18) � � ∶ � ∶ +all pairings n = odd � 1 2 3 4 n−2 n−1 n � In words, the right hand side is the sum over all possible contractions in the normal ordered product ∶ 1 � n ∶ (including the term with no contractions). The proof is by induction. We have already demonstrated this for n = 2, 3. Let W ( 1 , . . . , n ) stand for the right hand side of (4.18), and assume the theorem is valid for n − 1 fields. Then assuming x01 ≥ x02 ≥ � ≥ x0n , T( 1� n) = = = = 1T ( 1� n) 1W ( 2, . . . , n) (+) (−) ( 1 + 1 )W ( 2 , . . . , n ) (−) 1 W ( 2, . . . , n) + W ( 2, . . . , (+) n) 1 +[ (+) 1 , W ( 2, . . . , n )] 4.4. WICK’S THEOREM 71 The first two terms are normal ordered and contain all contractions that do not involve 1 . The last term involves the contractions of 1 with every field in every term in W ( 2 , . . . , n ), therefore all possible contractions. Hence the right hand side contains all possible contractions, which is W ( 1 , . . . , n ). 4.4.1 Combinatorics Let’s go back to our computation of S. We’d like to use Wick’s theorem to relate T [exp(i ∫ d4 x in (x)J(x))] to ∶exp(i ∫ d4 x in (x)J(x))∶, and for this we need a little combinatorics. To make the notation more compact we will continue dropping the “in” label on the “in” fields for the rest of this section. Now, T [exp(i ∫ d4 x in (x)J(x))] is a sum of terms of the form n in d4 xi J1 �Jn T ( � � n! i=1 1� n) = n in d4 xi J1 �Jn W ( � � n! i=1 1, . . . , Consider the term on the right hand side with one contraction, n 1 4 � � d xi J1 �Jn � ∶ n! i=1 pairs (i,j) = n) (4.19) 1� i� j � n∶ n−2 Npairs 4 � � d xi J1 �Jn−2 ∶ n! i=1 1 � n−2 ∶ � d4 x d4 x′ JJ ′ ′ where Npairs is the number of contractions in the sum, which is the same as the number of pairs (i, j) in the list 1, . . . , n. That is, the number of ways of choosing two elements of a list of n objects: Let Then n! n Npairs = � � = 2 2!(n − 2)! ⇣ = � d4 x d4 yJ(x)J(y)�0�T ( (x) (y))�0� . 1 pairing terms = n−2 −⇣ in−2 4 � � d xi J1 �Jn−2 ∶ 2 (n − 2)! i=1 1 � n−2 ∶ (4.20) . We want to repeat this calculation for the rest of the terms on the right hand side of (4.19). For this we need to count the number of terms for a given number of n contractions. Now, k contractions involve 2k fields. Now, there are � � ways of 2k picking 2k fields out of n, and we need to determine how many distinct contractions 72 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES one can make among them. We can figure this out by inspecting a few simple cases. For k = 1 12 and for k = 2 1234 1234 → one contraction → 3 contractions. 1234 Instead of continuing in this explicit way, we analyze the k = 3 case using inductive reasoning: 12 × (k = 2) 13 × (k = 2) � 16 × (k = 2) → 5 × 3 contractions. By induction the arbitrary k case has (2k − 1)!! pairings: �(2k − 1)-pairings:1j� × �(k − 1)-case: (2k − 3)!!� = (2k − 1)!! So the terms with k contractions give 1 n! (2k − 1)!! in−2k � n! (n − 2k)!(2k)! ��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� 1 1 k (n − 2k)! 2 k! = n−2k 4 � d xi J1 �Jn−2k ∶ i=1 m (−⇣�2)k im 4 � � d xi J1 �Jm ∶ k! m! i=1 1� m∶ 2 with m = n − 2k. Aha! We recognize this as a term in an exponential expansion. Considering all the terms in he expansion of T [exp(i ∫ d4 x in (x)J(x))], the coefficient of m ∞ 1 im 1 4 k m! ∫ ∏i=1 d xi J1 �Jm ∶ 1 � m ∶ (fixed m) is ∑k=0 k! (−⇣�2) = exp(− 2 ⇣), and is independent of m, so it factors out. We are left with ∞ m im 4 � � d xi J1 �Jm ∶ m=0 m! i=1 e−⇣�2 � That is S = T �exp �i � d4 x 1� m∶ in (x)J(x)�� k 4 4 ′ ′ ′ � � d x d x JJ ��� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � �� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � � (−1)k ⇣ k 1 � n−2k ∶ �i = e−⇣�2 ∶exp �i � d4 x J(x) (x)�∶ = e−⇣�2 ∶exp �i � d4 x J(x) (x)�∶ This will equal our previous expression for S in (4.17) if ⇣ = ⇠. To answer this we need to know more about the T-ordered product in the definition of ⇣ in (4.20). 4.5. SCALAR FIELD PROPAGATOR 4.5 73 Scalar Field Propagator Let F (x, y) ≡ �0�T ( (x) (y))�0� where (x) is a real, scalar, free field satisfying the Klein-Gordon equation. This is the quantity we need for the computation fo ⇣, but it is also important for several other reasons, so we spend some time investigating it. First of all, i F (x, y) is a Green’s function for the Klein-Gordon equation; see (4.3). To verify this claim compute directly. Taking @µ to be the derivative with respect to xµ keeping y µ fixed and using (@ 2 + m2 ) (x) = 0 we have (@ 2 + m2 ) F (x, y) = (@ 2 + m2 ) �✓(x0 − y 0 )�0� (x) (y)�0� + ✓(y 0 − x0 )�0� (y) (x)�0�� = @2 @2 0 0 ✓(x − y )�0� (x) (y)�0� + ✓(y 0 − x0 )�0� (y) (x)�0� @x02 @x02 @ @ (x) @ @ (x) + 2 0 ✓(x0 − y 0 )�0� (y)�0� + 2 0 ✓(y 0 − x0 )�0� (y) �0� 0 @x @x @x @x0 Now using d✓(x)�dx = (x) and @0 (x) = ⇡(x) the last line above is 2 (x0 − y 0 )�0�[⇡(x), (y)]�0� = −2i The line above that gives (4) (x − y) @ (x0 − y 0 )�0�[ (x), (y)]�0� = − (x0 − y 0 )�0�[⇡(x), (y)]�0� = i @x0 (4) Combining these we have (@ 2 + m2 ) F (x, y) = −i (4) (x − y) (x − y) As we saw in Sec. 4.1 Green functions are not unique, since one can always add solutions to the homogenous Klein-Gordon equation to obtain a new Green’s function. We must have d4 k ik⋅(x−y) 1 i F (x, y) = −i � e (2⇡)4 k 2 − m2 with some prescription for the contour of integration. Note that F (x, y) = F (x− y) depends only on the di↵erence x − y, which is as expected from homogeneity pof space-time. Recall that [ (+) (x), (−) (y)] = �0�T (x) (y)�0� for x0 > y 0 . More generally �0�T (x) (y)�0� = ✓(x0 − y 0 )[ Recall also that [ (+) (x), (−) (+) (x), (−) (y)] + ✓(y 0 − x0 )[ (+) 0 −y 0 )+ik � ⋅(� x −� y) (y)] = � (dk)e−iEk� (x (y), (−) (x)] 74 CHAPTER 4. INTERACTION WITH EXTERNAL SOURCES Without loss of generality and to simplify notation we set y µ = 0. We have �0�T (x) (0)�0� = ✓(x0 ) � (dk)e−iEk� x 0 +ik � ⋅� x + ✓(−x0 ) � (dk)eiEk� x 0 −ik � ⋅� x Comparing with Eq. (4.6) we see this is precisely −iGF (x), which was obtained by taking a contour that goes below −Ek� and above Ek� , Im(z) −Ek� + i✏ × × k 0 = Re(z) Ek� − i✏ So we have �0�T (x) (y)�0� = � d4 k −ik⋅(x−y) i e 4 2 (2⇡) k − m2 + i✏ (4.21) This will be of much use later. We will refer to this as the two-point function of the KG field, and the Fourier transform, 1�(k 2 − m2 + i✏), as the KG propagator. We can finally return to the question of relating ⇣ to ⇠: ⇣ = � d4 x d4 y J(x)J(y) �0�T (x) (y)�0� d4 k1 d4 k2 ik1 ⋅x ik2 ⋅y ̃ d4 p −ip⋅(x−y) i ̃ e e J(k ) J(k ) e 1 2 � (2⇡)4 (2⇡)4 (2⇡)4 p2 − m2 + i✏ d4 p ̃ ̃ i =� J(p)J(−p) 2 4 (2⇡) p − m2 + i✏ = � d4 x d 4 y � ̃ J(−p) ̃ ̃ 2 vanishes as �p0 � → ∞, Perform the integral over p0 , assuming J(p) = �J(p)� which is justified by our assumption that the source is localized in space-time. Closing the contour on the upper half of the complex p0 plane we pick the pole at −Ek� so that ⇣ = 2⇡i � d3 p ̃ 2 i ̃ 2 =⇠. �J(p)� = � (dk)�J(p)� (2⇡)4 −2Ek�